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Gravitation - Kepler and Newton revision notes and calculations.

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Kepler (1571-1630) has studied for many years the records of observation on planets and summarised three laws.

Kepler’s Law

  1. Each planet moves in an ellipse which has the sun at one focus
  2. The line joining the sun and the moving planet sweeps out equal area in equal time
  3. the square of the time of revolution of any planet (i.e. T) about the sun is proportional to the cube of the planets’ mean distance from the sun

I.e.   is a constant


Interpretation form Kepler’s laws

  1. Kepler’s second law :

        The area swept out in a very short time interval (Δt), neglecting the    

        small triangular region is A


         The rate of area swept =

         Hence it is a constant.  

         Compare this equation with the angular momentum

                                          = Constant

         This law is in fact an evidence of conservation of angular momentum.

  1. Kepler’s third law

       About 1666, Newton investigated the motion of the moon, and thought that it was the force of gravity to pull the moon and keep it in its orbit

NB: Time between full moon is 29.5 days but this due to the earth also moving round the sun, the moon is therefore to travel a bit longer

          At the earth surface g=9.81m . This is due to the fact that we are nearer the earth centre than the moon (about 1:60)

...read more.


             m= mg - m

             = g -    (< g )

  1. At latitude θ

        The body describes a circle of radius r centre



  1. Note that  mCosθ is the resultant force of mg and m.

    The direction of (or T) is not exactly towards the centre of the earth    

    except  at the poles and theequator

2) Practical Values


                   Latitude                          / m






Variation of g with heightimage02.png

(i)  r


➔                ➔      

Let        r = + h


If   h << then  

(ii)  r <  (We assume that the earth is a sphere of uniform density)

       Consider a small mass placed at A, the spherical shell does not produce

       and gravitational field on the point inside it.


     Since we assume that the earth has uniform densityimage03.png


∝ r

Graphical representationimage06.png


Practical data


          Altitude/ h(m)                              /(m)

              0                                                  9.81

          1000                                                9.80

          5                                           8.53

          1                                           7.41(Parking Orbit)

       3.8                                           0.0027 (Radius of moon orbit)


Satellite orbit

       The satellite can state in a fixed orbit with a specified radius, r, and

       corresponding velocity v by F = ma

                  --------------- (1)


                ---------------- (2)

          (2) →  (1):


For a satellite closed to the earth, stay at a height of about 100 –200km  

       Then  r

Parking Orbitimage08.png

        The earth is rotating, at this orbit; the satellite can stay constantly  

        Overhead of the earth

                                Angular velocity of the earth

                    F = ma  


                            h =


...read more.


          radius R. Determine the gravitational potential at a point distance r        

         from centre, where r < R.


    When  r < R        

➔        ➔    

      When r = R  


E.g.7 The masses of the earth and the moon are respectively  

           and  and they are separated by a    


  1.  Ignore the motions the earth and the moon , sketch the  

      gravitational potential V along the line joining them.

  1. Find the point where the net gravitational field strength (g) is  



  1. Choose the earth to be the origin

            V = Potential of earth + Potential of moon = +

      Near the earth, V is dominated by the field of the earth and near the  

        moon, V is dominated by the field of the  moon.


E.g.8 Some satellite, with the necessary electronic equipment inside, rises  

          vertically from the equator when it is fired. At a particular height

            the satellite is given a horizontal momentum by firing rocket on its  

          surface and the satellite then turns into the required orbit.

          A satellite is to be put into orbit 500km above the earth’s surface.

         If it’s vertical velocity after launching is 2000 at this height.

          Calculate the impulse required to put the satellite directly into    

Orbit, if its mass is 50kg.    (g = 10   Radius of earth =6400km)


      Suppose u is the velocity required for the orbit, radius R.  Then

      Force on the satellite =

                           u = 7700

     Let the impulse P required has the relation shown such that



...read more.

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